Question

The slope of tangent to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (2, –1) is ______.

विकल्प

• `22/7`

• `6/7`

• `(-6)/7`

• – 6

Answer 1

The slope of tangent to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (2, –1) is `6/7`.

Explanation:

The given curve is x = t² + 3t – 8 and y = 2t² – 2t – 5

`"dx"/"dt"` = 2t + 3 and `"dy"/"dt"` = 4t – 2

∴ `"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`

= `(4"t" - 2)/(2"t" + 3)`

Now (2, – 1) lies on the curve

∴ 2 = t² + 3t – 8

⇒ t² + 3t – 10 = 0

⇒ t² + 5t –2t – 10 = 0

⇒ t(t + 5) – 2(t + 5) = 0

⇒ (t + 5)(t – 2) = 0

∴ t = 2, t = – 5 and – 1 = 2t² – 2t – 5

⇒ 2t² – 2t – 4 = 0

⇒ t² – t – 2 = 0

⇒ t² – 2t + t – 2 = 0

⇒ t(t – 2) + 1(t – 2) = 0

⇒ (t + 1)(t – 2) = 0

⇒ t = – 1 and t = 2

So t = 2 is common value

∴ Slope `"dy"/("dx"_(x = 2)) = (4 xx 2 - 2)/(2 xx 2 + 3) = 6/7`